Hey people.
I have a sequence {b_n} which is defined like that :
b_n = 1+(1/3)+...+(1/1(2n-1)) , i.e b_n = sum (1/(2k-1)) from k=1 to n

I need to prove that lim (b_n) as n->inf = inf
I think I should use the squeeze theorem , but this sequence approaches infinity really slow, so I couldn't find a sequence {a_n} which approaches infinity and b_n >= a_n for almost every n...

Any clues anyone?
Thanks!

Mar 9th 2010, 05:08 AM

Prove It

Quote:

Originally Posted by Gok2

Hey people.
I have a sequence {b_n} which is defined like that :
b_n = 1+(1/3)+...+(1/1(2n-1)) , i.e b_n = sum (1/(2k-1)) from k=1 to n

I need to prove that lim (b_n) as n->inf = inf
I think I should use the squeeze theorem , but this sequence approaches infinity really slow, so I couldn't find a sequence {a_n} which approaches infinity and b_n >= a_n for almost every n...

Any clues anyone?
Thanks!

What you have posted is not a sequence, but a series.

Are you trying to evaluate the term of the series, or are you trying to show that the series is divergent?

Mar 9th 2010, 05:37 AM

Archie Meade

Quote:

Originally Posted by Gok2

Hey people.
I have a sequence {b_n} which is defined like that :
b_n = 1+(1/3)+...+(1/1(2n-1)) , i.e b_n = sum (1/(2k-1)) from k=1 to n

I need to prove that lim (b_n) as n->inf = inf
I think I should use the squeeze theorem , but this sequence approaches infinity really slow, so I couldn't find a sequence {a_n} which approaches infinity and b_n >= a_n for almost every n...